I have this statement:
The probability that it will rain tomorrow is $0.25$ and the
probability of cloudy dawn is $0.3$.What is the probability that tomorrow it will not rain nor the dawn be cloudy,
if you know that these two events are independent?
My attempt was:
Let $L$ the probability of rain, let $C$ the probability that it is a cloudy day.
First, $P(L \cap C) = P(L)\cdot P(C)$ since they are independent.
I need the probability of $P(L)^c \cap P(C)^c$, i.e that it does not rain and that at the same time, it is not a cloudy day. According to Morgan's law it is equal to $P(L \cup C)^c$. Therefore the probability is equal to $1 – \underbrace{(\frac{1}{4} + \frac{3}{10} – \frac{1}{4} \cdot \frac{3}{10})}_{P(L \cup C)} = 0.525$
But according to the guide, the correct answer must be $0.925$. What is wrong with my development? Thanks in advance.
The explanation of the guide was:
If the events are independent, then it rains and dawns cloudy has
probability $0,25 ยท 0,30 = 0,075$ , then the probability that it will
not be cloudy or new equals $1 – 0.075 = 0.925$
Best Answer
As commented on already, your derivation is correct.
The book's error is that it only finds the probability it will not be simultaneously rainy and cloudy (i.e. it gives $1 - P(L \cap C)$). It does not account for each event happening individually, which would be $1 - P(L \cup C)$ as you have noticed.